copyright © 2015, 2008, 2011 pearson education, inc. section 6.7, slide 1 chapter 6 polynomial...

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 1

Chapter 6Polynomial Functions


6.7 Using Factoring to Solve Polynomial

Equations


Zero Factor Property

Let A and B be real numbers.

If AB = 0, then A = 0 or B = 0.

In words, if the product of two numbers is zero, then at least one of the numbers must be zero.


Example: Solving a Quadratic Equation

Solve (x – 5)(x + 2) = 0.


Solution

2) 05( )(xx

0 or 5 02xx 5 or 2x x


Solution

Check that both 5 and –2 satisfy the original equation:

So, the solutions are 5 and –2.



Solve w2 – 2w – 8 = 0.


Solution

4) 02( )(ww

0 or 2 04ww 2 or 4w w

2 2 8 0w w


Check that both –2 and 4 satisfy the original equation:

So, the solutions are –2 and 4.

Solution


Example: Finding x-Intercepts of the Graph of a Quadratic Function

Find the x-intercepts of the graph of

f(x) = x2 – 7x + 10


Solution

To find the x-intercepts, substitute 0 for f(x) and solve for x:

So, the x-intercepts are (2, 0) and (5, 0).

2 70 10x x 0 ( )( )5 2x x

0 or 5 02xx 5 or 2x x


Solution

Use “zero” on a graphing calculator to verify our work.


Connection Between x-Intercepts and Solutions

Let f be a function. If k is a real-number solution of the equation f(x) = 0, then (k, 0) is an x-intercept of the graph of the function f. Also, if (k, 0) is an x-intercept of the graph of f, then k is a solution of f(x) = 0.



Solve 2x2 – 8x = 5x – 20.


Solution

22 8 5 20x x x 22 13 20 0x x

( )( )5 02 4x x

0 or 5 02 4xx 2 5 or 4x x

5 or 4

2x x


Solution

To verify that these are solutions to the equation, use a graphing calculator table to check that for each input, the outputs for y = 2x2 – 8x is equal to the output for y = 5x – 20.



Solve (x + 2)(x – 4) = 7.


Solution

Although the left-hand side of the equation is factored, the right-hand side is not zero. So first, find the product of the left-hand side:

( 2)( 4) 7x x 2 2 8 7x x

2 2 15 0x x ( 5)( 3) 0x x

5 0 or 3 0x x 5 or 3x x


Solution

Therefore, the solutions are –3 and 5. To verify the work, enter the equations y = (x + 2)(x – 4) and y = 7.


Solution

Use “intersect” to find the intersection points (–3, 7) and (5, 7). The x-coordinates of these points are the solutions of the original equation.


Example: Solving a Quadratic Equation That Contains Fractions

Solve 21 1 5.

2 3 6t t


Solution

To clear the equation of fractions, multiply both sides by the LCD, 6:

21 1 52 3 6

t t

26 61 1 5

66

2 3t t

23 2 5t t


Solution

23 2 5t t 23 5 2 0t t

( )( )2 03 1t t

0 or 2 03 1tt 3 2 or 1t t

2 or 1

3t t


Example: Finding an Input and an Output of a Quadratic Function

Let f(x) = x2 – 3x – 23.

1. Find f(5). 2. Find x when f(x) = 5.


Solution

1. f(5) = 52 – 3(5) – 23 = 25 – 15 – 23 = –13

2. Substitute 5 for f(x) in the equation:2 35 23x x 20 3 28x x

0 ( )( )7 4x x 0 or 7 04xx

7 or 4x x

Let f(x) = x2 – 3x – 23.


Solution

2. Verify that f(7) = 5 and f(–4) = 5 by hand.

Or, verify the work in Problems 1 and 2 using a graphing calculator table.


Solving Quadratic or Cubic Equations by Factoring

If an equation can be solved by factoring, we solve it by the following steps:

1. Write the equation so one side of the equation is equal to zero.2. Factor the nonzero side of the equation.3. Apply the zero factor property.4. Solve each equation that results from applying the zero factor property.


Example: Finding x-Intercepts of the Graph of a Cubic Function

Find the x-intercepts of the graph of

f(x) = x3 – 5x2 – 4x + 20


Solution

Substitute 0 for f(x) and solve for x:3 25 4 020x x x

2 4( 5) ( ) 05x x x

2 54 0( )xx

2( )( )( ) 052 xx x 20 or 0 or 052 xx x

2 or 2 or 5x x x

So, the x-intercepts are (–2, 0), (2, 0), and (5, 0).


Solution

Use “zero” on a graphing calculator to verify our work.


Zero Factor Property

Warning

The expression x2(x – 5) – 4(x – 5) = 0 is not factored, because it is a difference, not a product.

Only after we factor the left side of the equation can we apply the zero factor property


Cubic FunctionsThe solutions set of a cubic equation in one variable may contain one, two, or three real numbers (one, two, or three x-intercepts).


Example: Using Graphing to Solve an Equation in One Variable

Use graphing to solve x2 – x – 7 = –x2.


Solution

Use “intersect” on a graphing calculator to find the solutions of the system

y = x2 – x – 7 y = –x2


Solution

The approximate solutions of the system are (–1.64, –2.68) and (2.14, –4.57).

The x-coordinates of these ordered pairs are the approximate solutions of the equation.


Quadratic Model

Definition

A quadratic model is a quadratic function, or its graph, that describes the relationship between two quantities in an authentic situation.


Example: Modeling with a Quadratic Function

Annual revenues of restaurants are shown in the table for various years. Let f(t) be the annual revenue (in billions of dollars) of restaurants at t years since 1970. A model of the situation is

21( ) 5 54.

5f t t t


Example: Modeling with a Quadratic Function

1. Use a graphing calculator to draw the graph of the model and, in the same viewing window, the scattergram of the data. Does the model fit the data well?

2. Predict the revenue in 2018.3. Predict when the annual revenue was $84 billion.


Solution

1. The graph of the model and the scattergram of the data show that the model appears to fit the data quite well.


Solution

2. To predict the revenue in 2018, find f(48):

The revenue in 2018 will be $754.8 billion, according to the model.

21( ) 548 48 48 54 754.8

5f


Solution3. To estimate when the annual revenue was $84

billion, substitute 84 for f(t) and solve for t:

281

5 545

4 t t

210 5 30

5t t

20 25 150t t


Solution20 25 150t t 3.

The inputs –30 and 5 represent the years 1940 and 1975, respectively. The estimate of 1940 is an example of model breakdown, as a little research would show the revenue in 1940 was much less than $84 billion. Therefore, we estimate that it was 1975 when the revenue was $84 billion.

0 ( ) 3 )5 0(tt

0 or 35 00 tt 5 or 30t t


Solution

Use a graphing calculator table to verify our work in Problems 2 and 3.


Area of Rectangular Objects

The area A of a rectangle is given by the formula A = LW, where L is the length of the rectangle and W is the width.


Example: Solving an Area Problem

A person has a rectangular garden with a width of 9 feet and a length of 12 feet. She plans to place mulch outside of the garden to form a border of uniform width. If she has just enough mulch to cover 100 square feet of land, determine the width of the border.


Solution

Step 1: Define each variable. Let x be the width (in feet) of the mulch border.


Solution

Step 2: Write an equation in one variable. For the outer rectangle (garden and border), then length is 2x + 12 and the width is 2x + 9. So the area is:


Solution

Step 2: The area of the garden is 12 ∙ 9 = 108 square feet, and the area of the border is given as 100 square feet. The area of the garden plus the area of the border is equal to the area of the outer rectangle:


Solution

Step 3: Solve the equation.

208 (2 12)(2 9)x x 2208 4 18 24 108x x x 2208 4 42 108x x 20 4 42 100x x


Solution

20 4 42 100x x Step 3:

20 2 2 21 50x x

0 2(2 25)( 2)x x 2 25 0 or 2 0x x

2 25 or 2x x 25

or 22

x x


Solution

Step 4: Describe each result. For the result25

,2

x the border width is negative. Model breakdown has occurred, because a width must be positive. The border width is 2 feet.


Solution

Step 5: Check. If the border width is 2 feet, then the outer rectangle has a width of 13 feet and a length of 16 feet. Therefore, the total area of the outer rectangle is 13 ∙16 = 208 square feet. This checks with our calculation near the beginning of our work.

copyright © 2015, 2008, 2011 pearson education, inc. section 6.7, slide 1 chapter 6 polynomial...

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